Download PDF by David Olive: Statistical Theory and Inference

By David Olive

ISBN-10: 3319049712

ISBN-13: 9783319049717

This article is for a one semester graduate path in statistical idea and covers minimum and whole enough statistics, greatest chance estimators, approach to moments, bias and suggest sq. errors, uniform minimal variance estimators and the Cramer-Rao decrease certain, an creation to giant pattern concept, chance ratio exams and uniformly strongest exams and the Neyman Pearson Lemma. an enormous target of this article is to make those subject matters even more obtainable to scholars through the use of the idea of exponential families.

Exponential households, indicator capabilities and the aid of the distribution are used during the textual content to simplify the idea. greater than 50 "brand name" distributions are used to demonstrate the speculation with many examples of exponential households, greatest chance estimators and uniformly minimal variance independent estimators. there are lots of homework issues of over 30 pages of strategies.

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Extra resources for Statistical Theory and Inference

Example text

A) Show that f (y1 , y2 ) is a pdf. b) Find the marginal pdf of Y1 . Include the support. c) Find the marginal pdf of Y2 . Include the support. d) Find the conditional pdf fY1 |Y2 =y2 (y1 |y2 ). Include the support. e) Find the conditional pdf fY2 |Y1 =y1 (y2 |y1 ). Include the support. Solution. 1. The support is the region of integration Ω which is the triangle with vertices (0,0), (0,1), and (1,1). This triangle is bounded by the lines y1 = 0, y2 = 1, and y2 = y1 . The latter line can also be written as y1 = y2 .

Yik given any subset of the remaining random variables Y j1 = y j1 , . . ,Y jm = y jm is equal to the marginal: f (yi1 , . . , yik |y j1 , . . , y1 jm ) = f (yi1 , . . , yik ) = ∏kj=1 fYi j (yi j ) if f (y j1 , . . , y jm ) > 0. Proof. The proof for a joint pdf is given below. For a joint pmf, replace the integrals by appropriate sums. a) The marginal ∞ f (yi1 , . . , yik ) = −∞ ··· ∞ −∞ k = k ∞ ∏ j=k+1 −∞ j=1 dyik+1 · · · dyin j=1 n ∏ fYi j (yi j ) = n ∏ fYi j (yi j ) fYi j (yi j ) dyi j k fYi j (yi j ) (1)n−k = ∏ fYi j (yi j ).

4. In applications the support is often rectangular. For n = 2 the support is a cross product if Y = Y1 × Y2 = {(y1 , y2 ) : y1 ∈ Y1 and y2 ∈ Y2 } where Yi is the support of Yi . The support is rectangular if Y1 and Y2 are intervals. For example, if Y = {(y1 , y2 ) : a < y1 < ∞ and c ≤ y2 ≤ d}, then Y1 = (a, ∞) and Y2 = [c, d]. For a joint pmf, the support is rectangular if the grid of points where f (y1 , y2 ) > 0 is rectangular. 2 Expectation, Covariance, and Independence 35 2 4 y2 6 8 Cross Product of (1,2,3,4,9) with (1,3,4,5,9) 2 4 6 8 y1 Fig.

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Statistical Theory and Inference by David Olive


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